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Grade 11 CAPS Mathematics Video Series Circle Geometry I
Grade 11 CAPS
Mathematics
Video Series
Circle
Geometry I
In this Video we will :
and theorems linked to
and
related .
Investigate prove
perpendicular bisectors of chords
solve ride Lessrs on 1
Lessons linked to this Video
and theorems linked
to and and
related .
Investigate prove
inscribed central angles
solve rid Lesers son 2
Grade11 CAPS
Mathematics
Video Series
Lesson 1
Theorems
On
Perpendicular
Bisectors of
Chords
In this lesson we will :
Discuss the features of . Geometric Axiomatic System
Outcomes for Lesson 1
Recap the linked to circles. terminology
Investigate and prove that :
(
Segment from centre of circle perpendicular
to chord bisects chord Theor. em 1)
Investigate and prove that :
(
Segment from centre of a circle to midpoint
of chord is perpendicular to the Theo ch reord. m 2)
Investigate and prove that :
(
Perpendicular bisector of chord passes
through the centre of a circ Theorele. m 3)
Solve riders related to Theorems 1, 2 and 3.
An Axiomatic System consists of some:
(Accepted unproved statements or proved theorems)
(proved mathematical statements)
Undefined terms
Defined terms
Axioms
Theorems
: Point, line (segment; ray), angle, triangle,
exterior and interior angles of triangle
A straight angle measures
sExample from Geometry :
Undefined terms
Defined terms :
1 2
1 2
180
Axiom 1: corresponding angles are equal
Axiom 2: alternate angles are equal
Theorem 1: The sum of the interior angles of a triangle
is equal
Axioms :
Theorems : to 180
Theorem 2: The exterior angle of a triangle is equal to the
sum of the two interior opposite angles
Features of a Geometric Axiomatic System
180
Draw with
Draw
ABC
A B C
BD C BD
CE BA
Given :
Aim is to prove that :
Construction :
:
To avoid confusion
Axiom 2: alternate
A B C
A B ACB
ACE B ACB A ACE
Proof
Prove that : Given then 180ABC A B C
Axiom 1: corresponding ACE DCE ACB B DCE
180 Definition: is a straight angle BCD BCD
Once proved this theorem can be accepted as an additional axiom.
Proof of the second theorem, namely that ,
is left as an exercise.
ACD A B
A proof is a sequence of logical steps
A sound reason needs to be given
for each of the logical steps
Note :
Given a circle with centre
Refer to this as
O
O
Circle Terminology (Undefined terms)
The , , is the distance from the centre of
the circle to any point on the circumference.
rradius
The is a special chord that passes through
the centre of the circle.
diameter
A is the part of the circle that is cut off by a chord.
A chord divides the circle into two segments.
segment
An is a part of the circumference of a circle.arc
A is the line segment joining the ends of an arc.
Note: Chord can be linked to two arcs.
Or a chord is a line segment connecting two points on .
chord
Segment from centre of a circle
and perpendicular to a chord bisects the chord.
Investigation :
chord OM AB AM BM Conjecture 1 :
Many more investigations
possible by means of
GeoGebra.
Suggested Conclusion :
Any with chord
Draw and
O OM AB
AM BM
AO BO
Given :
Aim is to prove that :
Construction :
2 2 2
Assume the Theorem of Pythagoras.
with 90ABC B AC AB BC
Axiom 1 :
2 2 2
:
Theorem of PythagorasAM AO OM
Proof
2 2 BO OM AO BO r
2 Theorem of Pythagoras BM
AM BM
Segment from centre of a circle
and perpendicular to a chord bisects the chord.
Theorem :
chord OM AB AM BM Theorem 1 :
Line segment from centre of circle to
midpoint of chord is perpendicular to the chord.
Investigation :
chord AM MB OM AB Conjecture 2 :
Many more investigations possible by means of GeoGebra.
Suggested Conclusion :
Any with the midpoint of chord
Draw , and
O M AB
OM AB
AO BO OM
Given :
Aim is to prove that :
Construction :
Assume two 's are congruent
Three sides of one triangle are equal
to three sides of the other triangle
Axiom 2 :
: In 's and ,
radii
common
given
OAM OBM
AO BO
OM OM
AM BM
Proof
, ,
2 1 90 180
OAM OBM s s s
AMB
OM AB
Theorem 2 is the converse of Theorem 1.
chord OM AB AM BM
Line segment from centre of circle to
midpoint of chord is perpendicular to the chord.
Theorem :
chord AM MB OM AB Theorem 2 :
Perpendicular bisector of a chord
passes through the centre of a circle.
Investigation :
MN AB AM BM O MN Conjecture 3 :
Suggested Conclusion :
Any with the midpoint of chord
Draw , and
O M AB
AO BO
AO BO OM
Given :
Aim is to prove that :
Construction :
is a perpendicular bisector of chord
OM AB
O OM NM
Perpendicular bisector of a chord
passes through the centre of a circle.
Theorem :
: In 's and ,
radii
common
given
OAM OBM
AO BO
OM OM
AM BM
Proof
, ,
2 1 90 180
OAM OBM s s s
AMB
OM AB
MN AB AM BM O MN Theorem 3 :
Constructing the centre of any given circle
Theorem 3 can be utilized to
construct the centre of any circle :
Given any
Draw perpendicular
bisectors to any two chords
Centre is the point of intersection
of these two perpendicular bisectors
Constructing the circumscribed circle about a given triangle
The point of concurrency of the
perpendicular bisectors of the
sides of a triangle is called the
of the triangle.circumcentre
Draw perpendicular bisectors
and of and
respectively.
With as centre and as
radius draw circumscribed circle.
DF EG AB BC
DF EG O
O OB
Method :
, ,
, ,
Thus and radius
will pass through , and .
AOD BOD s s AO OB
BOE COE s s OB OC
AO OB OC
O OB
A B C
Why does this method work?
The circle which passes through
the vertices of the triangle is the
or
of the triangle.
circumcircle circumscribed
circle
Find the value of in each of the following ridersx
2 2 2
2 2
42
10
Pyth
10 4 2 21
FGFH
FE EI r
x FE FH
x
Solution : Rider 2
2 2 2
2 2
bisects
42
(Pyth)
5 4 3
AD BC
BCBD
x AB BD
x
Solution : Rider 1
2 2 2
22 2
2 2
2
Pyth
2 6
4 4 36
4 40 10
JL JK r x
NJ JL LN x
JN KM
KJ NJ KN
x x
x x x
x x
Solution : Rider 3
PAUSE Video
• Do Tutorial 1
• Then View Solutions
Tutorial 1: Find the value of in each of the following ridersx
Tutorial 1: Rider 1: Suggested Solution
2 2 2
2 2 2
2 2
32
Pyth
3 4
3 4 5
BCBD
AB BD AD
x
x
Solution : Rider 1
Tutorial 1: Rider 2: Suggested Solution
2 2 2
2 2
2 2
: Pyth
10 8 6 82
EJ FE FJ EFJ
EJ FE FJ
FGFJ
Solution : Rider 2
2 2 2
2 2
2 2
: Pyth
10
10 5 5 3 5
2
EK EI KI EKI
EK EI KI
EI r
HIKI
5 3 6x JK EK EJ
Tutorial 1: Rider 3: Suggested Solution
122
12.52
NPNR
NONS
Solution : Rider 3
2 2 2
2 2
: Pyth
12 5 13
NM NR MR NRM
NM r
2 2 2
2 2
22
: Pyth
51 13 12.5
2
MS NM NS NSM
x NM NS
More riders related to perpendicular bisectors of chords
Make an appropriate sketch
Prove each by utilizing Theorems 1 to 3 and other axioms
For each of the two additional riders :
is the centre of two concentric circles.
Chord of the greater circle cuts the smaller circle at and .
Prove that .
O
AB C D
AC DB
Rider 4 :
Two circles with centres and intersect at and .
with is parallel to , with and on the two circles.
Prove that 2 .
M N A B
PQ A PQ MN P Q
PQ MN
Rider 5 :
Rider 4: Suggested Solution
is the centre of two concentric circles.
Chord of the greater circle cuts the smaller circle at and .
Prove that .
O
AB C D
AC DB
Rider 4 : Concentric circles are
circles with common radius
Draw OE CDConstruction :
:
bisects chord
bisects chord
AE BE OE AB
CE DE OE CD
Proof
AE CE BE DE
AC DB
Rider 5: Suggested Solution
Two circles with centres and intersect at and .
with is parallel to , with and on the two circles.
Prove that 2 .
M N A B
PQ A PQ MN P Q
PQ MN
Rider 5 :
Draw
and
MR AP
NS AQ
Construction :
:
bisects chord 2
bisects chord 2
APAR MR AP
AQAS NS AQ
Proof
m Opposite sides of MN RS RMNS
But MN RS AR AS
2 2 2 2
2
AP AQ AP AQ PQMN AR AS
PQ MN
22 2
is a chord of and with
is drawn such that 2 .
3Prove that .
2
AB M AC B AC
BC AB
BCMC MB
Rider 2 :
PAUSE Video
• Do Tutorial 2
• Then View Solutions
Tutorial 2: Riders linked to perpendicular bisectors of chords
Make an appropriate sketch
Prove each by utilizing Theorems 1 to 3 and other axioms
For each of the two riders :
Prove that equal chords are equidistant from the
centre of a circle.
Rider 1 :
Tutorial 2: Rider 1: Suggested Solution
Prove that equal chords are equidistant from the centre of a circle.
Rider 1 :
with
Draw
and
O AB DE
OC AB
OF DE
OC OF
Given :
Construction :
Aim is to prove that :
:
bisects chord and2
bisects chord 2
ABBC OC AB
DEFE OF DE
Proof
2 2
AB DEBC FE AB DE
2 2 2
2 2
2
2 2
: Pyth
: Pyth
OC r BC OCB
r FE BC FE
OF OFE
OC OF OC OF
r
r
Tutorial 2: Rider 2: Suggested Solution
22 2
is a chord of and with
is drawn such that 2 .
3Prove that .
2
AB M AC B AC
BC AB
BCMC MB
Rider 2 : Draw ,
, and .
MD AB
AM MB MC
Construction :
2 2 2
2 2 2
2 2 2 2
1
:
: Pyth
: Pyth
From 1 and 32
2
MC MD DC MDC
MD MB DB MDB
MC MB DB DC
Proof
bisects chord 2
1 2
2 2 4 24
ABDB MD AB
BC BC BCBC AB AB
and Obvious
5 From 4
4 45
DC DB BC
BC BCBC
2 2
2 2
2 22
22
22
Substituting 4 and 5 into 3 :
5
4 4
25
16 16
24
16
3
2
BC BCMC MB
BC BCMB
BCMB
BCMB
Grade11 CAPS
Mathematics
Video Series
Lesson 2
Inscribed and
Central Angle
Theorems
In this lesson we will :
Outcomes for Lesson 2
Investigate and prove that :
(On the sam
Angle subtended by an arc (or chord) at the
centre of a circle is double the size of the angle subtended by the same
arc at the circumfrence of the circle.
e side of the chord as the centre)
The central angle is double the inscribed angle subtended by the same chord.
(
Alternative for
Theorem 4)
mulation :
Solve riders related to Theorems 4 and 5.
Investigate and prove that :
Angles subtended by a chord at the
circumference of a circle, on the same side of the chord, are equal.
( Theorem 5)
An angle such as in the accompanying sketch,
whose vertex lies on a circle and
whose sides are chords of a circle,
is called an in the circle.
C
inscribed angle
Inscribed and central angle terminology
We also say that the or the chord
AB
ACB
major arc AB
subtends on the circle.
whose vertex is the centre of the circle,
is called a of the circle.
(Note sides of this angle are radii)
AOB
central angle
We say that the or the chord
AB
AOB
major arc AB
subtends at the centre.
The central angle of a circle is double
the inscribed angle subtended by the same chord.
Investigation :
with inscribed and central
2
O BAC BOC
BOC BAC
Conjecture :
for more!GeoGebra
:Suggested Conclusion
with inscribed and central
2
O BAC BOC
BOC BAC
Prove that : with inscribed and central
2
Draw with
O BAC BOC
BOC BAC
AD O AD
Given :
Aim is to prove that :
Construction :
:
1 2 ; radii of AO OB
Proof
s1 2 ext. sum of int. opp.
2 1
BOD
s
Similarly: 3 4
2 3
ext. sum of int. opp.
COD
see sketch
2 1 2 3 2 1 3 2
BOC BOD COD
BAC
The central angle of a circle is double
the inscribed angle subtended by the same chord.
Theaorem :
with inscribed
and central
2
O BAC
BOC
BOC BAC
Theorem :
Corollaries (Deductions) from Theorem 4
The angle in a semicircle is a right angle.Corollary 1 :
18090
2 2
BOCA
Chord are equal if they subtend
equal angles at points on a circle.
Proof will be discu ssed as an exer e s . ci
Corllary 2 :
Equal chords subtend equal inscribed angles
at points on the same circle.
Proof left as an exe cise. r
Corollary 3 :
Equal chords in the same circle subtend equal central angles.
Proof left as exercise.
Corollary 4 :
Simple Riders linked to Theorem 4
Determine the value of angles represented by Greek letters.
Rider 1
44
180 2 44 92 Sum of angles
46 2
ABC ACB AB AC
Solution :
Rider 2
180 8348.5 Reason?
2
8341.5 Reason?
2
Solution :
Rider 3
50 Reason?
130 Reason?
Solution :
Chords are equal if they subtend equal inscribed angles
Proof of Corollary 2 :
with
Join , , and to
O BAC EDF
BC EF
B C E F O
Given :
Aim is to prove that :
Construction :
:
2 and 2 central 2 inscribed
Given that
BOC A EOF D
BOC EOF A B
Proof
In and : proved
radii
radii
, ,
BOC EOF BOC EOF
OB OE
OC OF
BOC EOF s s
BC EF
Corollaries 3 and 4 may be
proved in the same way.
Proofs left as an exercise!
Tutorial 3: Central 2 inscribed
1 Determine the value of the angles represented by Greek letters.
2 and are two parallel chords of .
Prove that .
XY ST M
XT YS P
XMS XPS
3 is a diameter of and a chord.
The circle on as diameter cuts at .
Prove that . 2
AB O AC
AO AC M
BCMO
PAUSE Video
• Do Tutorial 3
• Then View Solutions
Tutorial 3: Problem 1: Rider 1: Suggested Solution
2 central 2 inscribed CAB D
46 2
2 46
2 46
46
r r
r r
r r
r
1 Determine the value of the angles represented by Greek letters.
Tutorial 3: Problem 1: Rider 2: Suggested Solution
180 180 180 98 82GEF GEN NEF
41 central 2 inscribed angle2
GEFGEF
41 central 2 inscribed angle2
GEFGEF
1 Determine the value of the angles represented by Greek letters.
Tutorial 3: Problem 1: Rider 3: Suggested Solution
Obtuse
central 2 inscribed angle2
11055
2
KIJ
Reflexive 360 obtuse
360 110 250
KIJ KIJ
Reflexive
central 2 inscribed angle2
250125
2
KIJ
1 Determine the value of the angles represented by Greek letters.
Tutorial 3: Problem 2: Suggested Solution
2 and are two parallel chords of .
Prove that .
XY ST M
XT YS P
XMS XPS
2 central 2 inscribed angleXMS T 1
2 central 2 inscribed angle
2 2
XMS Y
T Y T Y
2
alt. YXP T XY ST 3
ext. of
from 2 and 3
2 from 1
XPS Y YXP XYP
T T
T XMS
Tutorial 3: Problem 3: Suggested Solution
3 is a diameter of and a chord.
The circle on as diameter cuts at .
Prove that . 2
AB O AC
AO AC M
BCMO
90 angles in semicirclesAMO ACB
sin is a right-angled
and sin is a right-angled
OMA OMA
AO
CBA BCA
AB
OM CB AOOM CB
AO AB AB
2
AOCB AO BO
AO
2
BCOM
Angles subtended by a chord (or arc) of a circle,
on the same side of the chord, are equal.
Investigation :
Inscribed angles on the same side of a chord are equalConjecture :
for more!GeoGebra
:Suggested Conclusion
Angles on same chord on different
sides of chord are supplementary.
(More about this in next video)
Note :
with inscribed and inscribed
on the same side of chord
O BAC BDC
BC BAC BDC
Theorem 5 :
with inscribed and inscribed
angle on the same side of chord
Draw and
O BAC
BDC BC
BAC BDC
OB OC
Given :
Aim is to prove that :
Construction :
:
central 2 inscribed 2
BOCBDC
Proof
transitive property for equalityBDC BAC
and
central 2 inscribed 2
BOCBAC
Angles subtended by a chord (or arc) of a circle,
on the same side of the chord, are equal.
Theorem :
Simple Riders linked to Theorem 5
s
Rider 2
37 same chord
37 alternate
37 same chord
JG
HF
Solution :
Rider 1
48 angles on chord
35 angles on chord
BC
AD
Solution :
Rider 3: Give reasons
34 reason ?
180 104 reason ?
180 34 104 42
NPQ
NPQ
Solution :
42 reason ?NPM
1 Determine the value of the angles represented by Greek letters.
Tutorial 4: Angles subtended by a chord at a point on circle are equal
PAUSE Video
• Do Tutorial 4
• Then View Solutions
1 Determine the value of the angles represented by letters.
Tutorial 4: Rider 1: Suggested Solution
Determine the value of angles represented by letters in the rider.
Rider 1
40 angles on same chord
36 angles on same chord
a AD
b BC
Solution :
Tutorial 4: Rider 2: Suggested Solution
Rider 2
33 angles on same chord
33 alternate angles
33 angles on same chord
c HF
e c
d e JG
Solution :
Determine the value of angles represented by letters in the rider.
Tutorial 4: Rider 3: Suggested Solution
: Rider 3:
34 angles on same chord f LP
Solution
180 34 49 55 =42 sum of angles in h NMP
42 (Sum of inner angles of a )g h
49 angles on same chord j ML
34 42 76 ext. of sum of opp. int. anglesk f g
Determine the value of angles represented by letters in the rider.
End of the First Video on Circle Geometry
REMEMBER!
•Consult text-books for additional examples.
•Attempt as many as possible other similar examples
on your own.
•Compare your methods with those that were
discussed in this Video.
•Repeat this procedure until you are confident.
•Do not forget:
Practice makes perfect!
